Differential equation using Laplace transform struck on inverse Laplace I was solving this differential equation 
$$ty''+2y'+ty=\cos{t}$$
with initial condition $y(0)=1$.
I took Laplace on both sides and after simplifying got this differential equation in terms of $Y(s)$
$$(s^2-2)Y'(s)+4sY(s)=\frac{s}{s^2+1}+3$$
Solving this I got $$Y(s)=\frac{s^2- 3\log{(s^2+1)}+2s^3-12s}{2(s^2-2)^2}$$
What is the Laplace inverse of $Y(s)$?
 A: This is a linear non-homogeneous differential equation. It can be solved as
$$
y = y_h + y_p
$$
such that
$$
ty''_h+2y'_h+ty_h = 0\\
ty''_p+2y'_p+ty_p = \cos t
$$
To solve $y_h$ we make $y_h = \frac{e^{a t}}{t}$ and substituting
$$
(1+a^2)e^{a t} = 0 \Rightarrow y_h = \frac{C_1\sin(t)+C_2\cos(t)}{t}
$$
Now substituting $y_p = C_1\sin(t)+C_2\cos(t)$ into the particular equation we get
$y_p = \frac{1}{2}\sin(t)$ and finally
$$
y =  \frac{C_1\sin(t)+C_2\cos(t)}{t} + \frac{1}{2}\sin(t)
$$
A: Your Laplace equation isn't correct. Since you didn't show your work, I can't say exactly what was wrong.
Remember $\mathcal L \{ty(t)\} = -\frac{d}{ds}Y(s)$, so we have
$$ \mathcal L \{ty''\} = -\frac{d}{ds} (s^2Y - s - y'(0)) = -(2sY + s^2Y' - 1) $$
Therefore
$$ -(2sY + s^2Y' - 1) + 2(sY - 1) - Y' = \frac{s}{s^2+1} $$
$$ -(s^2+1)Y' = \frac{s}{s^2+1} + 1 $$
$$ -Y' = \frac{s}{(s^2+1)^2} + \frac{1}{s^2+1} $$
You can take the inverse transform of the above to obtain
$$ t y(t) = \frac12 t\sin t + \sin t $$
$$ y(t) = \frac12 \sin t + \frac{\sin t}{t} $$
